Part 4: Buffer action, preparation and calculations

Explain acidic and basic buffers and calculate acidic-buffer pH before and after small additions.

A buffer needs both members of a conjugate pair

A buffer resists large pH changes when small amounts of acid or base are added and approximately maintains pH on moderate dilution. It does not keep pH exactly fixed or absorb unlimited additions. An acidic buffer commonly contains a weak acid and a soluble salt supplying its conjugate base, such as ethanoic acid and sodium ethanoate.

Added H⁺ is consumed by A⁻ to form HA. Added OH⁻ reacts with HA to form A⁻ and water. Because both components are present in substantial amounts, their ratio changes only slightly for small additions. A weak acid alone lacks a large reserve of conjugate base to remove added acid.

Basic buffers also need a conjugate pair

An ammonia/ammonium buffer contains NH₃ and a salt supplying NH₄⁺. NH₃ removes added H⁺ by forming NH₄⁺. NH₄⁺ removes added OH⁻ by forming NH₃ and water. Write these reactions with correct charges rather than saying only “the equilibrium opposes the change”.

Buffers help maintain conditions for enzyme activity, analytical measurements and industrial processes. The carbonic-acid/hydrogencarbonate system contributes to blood pH regulation alongside respiration and other systems; it is not an isolated unlimited buffer.

Rearrange Ka and use the final ratio

From Ka = [H⁺][A⁻]/[HA], [H⁺] = Ka[HA]/[A⁻]. When both components are in the same solution volume, their mole ratio equals their concentration ratio. This cancellation is valid for the ratio, not for every expression involving concentration.

For a buffer containing 0.0240 mol HA and 0.0360 mol A⁻ with Ka = 1.80 × 10⁻⁵, [H⁺] = 1.80 × 10⁻⁵ × 0.0240/0.0360 = 1.20 × 10⁻⁵ mol dm⁻³. pH = 4.92. The logarithmic rearrangement gives the same result.

Two routes to a buffer

Mixing measured amounts of weak acid and its salt directly gives both components. If the salt is supplied as a mass, calculate moles using its actual formula, including water of crystallisation if present. For example, 1.64 g anhydrous sodium ethanoate with Mr = 82.0 supplies 0.0200 mol ethanoate.

Alternatively, partially neutralise the weak acid with a strong base. The base consumes HA and makes A⁻. Work out the new moles before using Ka. Complete neutralisation leaves no substantial HA reserve and is not the same buffer mixture.

Update both components after an addition

Start from the earlier 0.0240 mol HA/0.0360 mol A⁻ buffer. Add 0.00200 mol strong acid: A⁻ falls to 0.0340 mol and HA rises to 0.0260 mol. [H⁺] = 1.80 × 10⁻⁵ × 0.0260/0.0340, giving pH = 4.86.

In a separate identical original buffer, add 0.00200 mol NaOH instead. HA becomes 0.0220 mol and A⁻ becomes 0.0380 mol; pH = 4.98. These changes are small relative to the starting 4.92 because neither reservoir is close to exhaustion.

If an addition consumes essentially all of one component, stop using the buffer approximation. Identify the remaining solution: excess strong acid or alkali, weak acid alone, or a hydrolysing salt may control its pH. Negative component moles are a signal that the wrong calculation model is being used.

Choose a ratio and preserve capacity

For target pH equal to pKa, use approximately equal acid and conjugate-base concentrations. More generally n(A⁻)/n(HA) = 10^(target pH − pKa). If pKa = 4.80 and target pH = 5.10, the required ratio is about 2.00. For a total of 0.0600 mol buffer components, that means approximately 0.0400 mol A⁻ and 0.0200 mol HA.

Moderate dilution changes both concentrations by the same factor, so their ratio and approximate pH stay the same. However, a fixed volume of diluted buffer contains fewer moles and has lower capacity to neutralise added acid or base. Very extensive dilution can invalidate the assumptions. A buffer works most effectively when neither conjugate component is present in a very small amount.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources